Question

A meteoroid, heading straight for Earth, has a speed of 14.8 km/s relative to the center of Earth as it crosses our moon's orbit, a distance of 3.84 × 108 m from the earth's center. What is the meteroid's speed as it hits the earth

Answer 1

Answer:

The meteoroid's speed is 18.5 km/s

Explanation:

Given that,

Speed = 14.8 km/s

Distance $d= 3.84\times10^{8}$

We need to calculate the meteoroid's speed

The total initial energy

$E_{i}=K_{i}+U_{i}$

$E_{i}=\dfrac{1}{2}mv_{i}^2-\dfrac{GM_{e}m}{r}$

Where, m = mass of  meteoroid

G = gravitational constant

$M_{e}$=mass of earth

r = distance from earth center

Now, The meteoroid hits the earth then the distance of meteoroid from the earth 's center will be equal to the radius of earth

The total final energy

$E_{f}=K_{f}+U_{f}$

$E_{f}=\dfrac{1}{2}mv_{f}^2-\dfrac{GM_{e}m}{r_{e}}$

Where,

$r_{e}$=radius of earth

Using conservation of energy

$E_{i}=E_{j}$

Put the value of initial and final energy

$\dfrac{1}{2}mv_{i}^2-\dfrac{GM_{e}m}{r}=\dfrac{1}{2}mv_{f}^2-\dfrac{GM_{e}m}{r_{e}}$

$v_{f}^2=v_{i}^2+2GM_{e}(\dfrac{1}{r_{e}}-\dfrac{1}{r})$

Put the value in the equation

$v_{f}^2=(14.8\times10^{3})^2+2\times6.67\times10^{-11}\times5.97\times10^{24}(\dfrac{1}{6.37\times10^{6}}-\dfrac{1}{3.84\times10^{8}})$

$v_{f}=\sqrt{(14.8\times10^{3})^2+2\times6.67\times10^{-11}\times5.97\times10^{24}(\dfrac{1}{6.37\times10^{6}}-\dfrac{1}{3.84\times10^{8}})}$

$v_{f}=18492.95\ m/s$

$v_{f}=18.5\ km/s$

Hence, The meteoroid's speed is 18.5 km/s